Bounds on coefficients: univalent maps

Question

Suppose $f(z)=c_1z+c_2z^2+\cdots+c_nz^n$ is a univalent map on the unit disk.

You may assume $c_1=1$. All coefficients are complex.

Is there a sharp bound on the modulus of the last coefficient $c_n$?

How about the other coefficients?

Answer 1

There is something to say about $c_n$, if not much about the lower order coefficients.

Since $f(z)$ is a schlicht function, $f'(z)\neq0$ throughout the disk. That means, each root of the polynomial $f'(z)$ has modulus $\vert z\vert\geq1$. From $$f'(z)=1+2c_2z+\cdots+nc_nz^{n-1},$$ we know that the product of the roots is $\frac1{nc_n}$ and this quantity must satisfy $$\frac1{n\vert c_n\vert}\geq1.$$ In other words, $n\vert c_n\vert\leq1$ or $\vert c_n\vert\leq\frac1n$, which is indeed stronger than $\vert c_n\vert\leq n$.

Answer 2

Concerning the other coefficients $c_k$, Branan [1] showed that $(k+1)c_{k+1}=(n-k)\bar c_{n-k}$ is a necessary condition for univalence in $|z|<1$ if $c_1=1$ and $c_n=1/n$.

For univalent polynomials with real coefficients and such that $c_{n}=1/n$, Suffridge [2] has proved that the coefficients $c_{k}$ satisfy the inequalities $$|c_{k}|\leq |C_{k,1}|, $$ and for univalent polynomials with real coefficients and such that $c_{n}=-1/n$, $$ |c_{k}|\leq\begin{cases} |C_{k,1}|\quad(n\text{ even})\\ |C_{k,2}|\quad(n\text{ odd}), \end{cases} $$ where $$C_{k,j}=\frac{n-k+1}{n}\frac{\sin jk\pi/(n+1)}{\sin j\pi/(n+1)},\qquad k=2,\ldots,n,\quad j=1,\ldots,n.$$ For each $n=1,2,\ldots$, and $1\leq j\leq n$, the polynomials $$P(z)=\sum_{k=1}^n C_{k,j}z^k$$ are univalent in $|z|<1$, from which follows that the above inequalities are sharp.

[1] D. A. Brannan, Coefficient regions for univalent polynomials of small degree, Mathematika, 14 (1967), 165-169.

[2] T. J. Suffridge, On univalent polynomials. J. London Math. Soc. 44 (1969), 496-504.